Giúp mình hai câu này với

Câu 6

Ta có

$\sin \alpha = \dfrac{3}{5}$

Suy ra

$\cos \alpha = \sqrt{1 - \sin^2\alpha} = \sqrt{1 - \dfrac{9}{25}} = \pm \dfrac{4}{5}$

Do $\dfrac{\pi}{2} < \alpha < \pi$ nên $\cos \alpha < 0$

Vậy $\cos \alpha = -\dfrac{4}{5}$

Vậy 

$\tan \left( \alpha + \dfrac{\pi}{3} \right) = \dfrac{\sin \left( \alpha + \frac{\pi}{3} \right)}{\cos \left( \alpha + \frac{\pi}{3} \right)}$

$= \dfrac{\sin \alpha \cos\left( \frac{\pi}{3} \right) + \cos \alpha \sin \left( \frac{\pi}{3} \right)}{\cos \alpha \cos \left( \frac{\pi}{3} \right) - \sin \alpha \sin \left( \frac{\pi}{3} \right)}$

$= \dfrac{\frac{3}{5} . \frac{1}{2} -\frac{4}{5} . \frac{\sqrt{3}}{2} }{-\frac{4}{5}. \frac{1}{2} - \frac{3}{5}.\frac{\sqrt{3}}{2}}$

$= \dfrac{48 - 25\sqrt{3}}{11}$

Câu 7

Ta có

$\sin \alpha = \dfrac{1}{\sqrt{3}}$

Suy ra

$\cos \alpha = \sqrt{1 - \sin^2\alpha} = \sqrt{1 - \dfrac{1}{3}} = \pm \dfrac{\sqrt{6}}{3}$

Do $\dfrac{\pi}{2} > \alpha > 0$ nên $\cos \alpha > 0$

Vậy $\cos \alpha = \dfrac{\sqrt{6}}{3}$

Vậy 

\cos \left( \alpha + \dfrac{\pi}{3} \right) = \cos \alpha \cos \left( \dfrac{\pi}{3} \right)  - \sin \alpha \sin \left( \dfrac{\pi}{3} \right)$

$= \dfrac{\sqrt{6}}{3} . \dfrac{1}{2} - \dfrac{1}{\sqrt{3}} . \dfrac{\sqrt{3}}{2}$

$= \dfrac{1}{\sqrt{6}} - \dfrac{1}{2}$

Đáp án A.